Using Remainder and Factor Theorem, factorise the following polynomial :

Given,

$f(x) = 2x^3 + 5x^2 - 28x - 15$

Substituting $x = 3$ in $f(x)$, we get :

$$\begin{aligned}\Rightarrow f(3) = 2(3)^3 + 5(3)^2 - 28(3) - 15 \\[1em] \Rightarrow f(3) = 2 \times 27 + 5 \times 9 - 84 - 15 \\[1em] \Rightarrow f(3) = 54 + 45 - 84 - 15 \\[1em] \Rightarrow f(3) = 99 - 99 \\[1em] \Rightarrow f(3) = 0\end{aligned}$$

Since, the remainder is zero.

∴ $(x - 3)$ is a factor of $f(x)$.

Dividing $2x^3 + 5x^2 - 28x - 15$ by $(x - 3)$, we get :

$$\begin{array}{l} \phantom{x - 3\big)\quad } 2x^2 + 11x + 5 \\ x - 3\overline{\smash{\big)}\quad 2x^3 + 5x^2 - 28x - 15} \\ \phantom{+++(}\underline{\underset{-}{+}2x^3 \underset{+}{-}6x^2} \\ \phantom{x - 3\big)\quad }\phantom{2x^3 +} 11x^2 - 28x \\ \phantom{+++( }\phantom{2x^3 +} \underline{\underset{-}{+}11x^2 \underset{+}{-}33x} \\ \phantom{++++( }\phantom{2x^3 + 5x^2 -} 5x - 15 \\ \phantom{++++) }\phantom{2x^3 + 5x^2 -} \underline{\underset{-}{+}5x \underset{+}{-}15} \\ \phantom{x - 3\big)\quad }\phantom{2x^3 + 5x^2 - 28x} \times \end{array}$$

∴ $2x^3 + 5x^2 - 28x - 15 = (x - 3)(2x^2 + 11x + 5)$

Factorising $2x^2 + 11x + 5$, we get :

$$\begin{aligned}= (x - 3)[2x^2 + 10x + x + 5] \\[1em] = (x - 3)[2x(x + 5) + 1(x + 5)] \\[1em] = (x - 3)(x + 5)(2x + 1).\end{aligned}$$

Hence, $\bm{{2x^3} + 5x^2 − 28x − 15 = (x − 3)(x + 5)(2x + 1)}$.